What does "$\binom n k$" mean? \begin{align}
& \binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom n n = \\[8pt]
& \text{(A)}\,\,\, 2^{n-1} \qquad \text{(B)} \,\,\, {}^{2n}C_n \qquad \text{(C)} \,\,\,2^n \qquad \text{(D)}\,\,\,2^{n+1}
\end{align}
Why is there nothing inside the bracket (between the $n$ and the number)?
 A: Simply,
$$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$
A: The numbers inside the brackets don't represent fractions. They are known as binomial coefficients and that is how we write them. For example,
$$\binom{3}{2}=\frac{3!}{2!1!}$$
Very simply, the binomial coefficent $\binom{n}{k}$ (pronounced $n$ choose $k$) represents the number combinations that can be formed from $k$ items out of $n$ items.
A: $${n \choose k}$$ means "for a set of size $n$, how many subsets have size $k$" ?  For example, the set $\{a, b, c, d\}$ of size 4 has 6 subsets of size 2: $\{a, b\}, \{a, c\}, \{a, d\}, \{b, c\}, \{b, d\}, \{c, d\}$
, so
$${4 \choose 2} = 6$$

$$\begin{align}
& \binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom n n = \\[8pt]
& \text{(A)}\,\,\, 2^{n-1} \qquad \text{(B)} \,\,\, {}^{2n}C_n \qquad \text{(C)} \,\,\,2^n \qquad \text{(D)}\,\,\,2^{n+1}
\end{align}$$

This is asking "how many subsets of size 0 plus subsets of size 1 plus...." which can become the question "how many subsets are there (from a set of size n)?"
